Question

WILL MEDAL!!!!!!
1. Suppose that the population of deer in the state is 19,900 and is growing 3% each year. Predict the population after 10 years.
a. about 274,338 deer
b. about 26,744 deer
c. about 597,000 deer
d. about 204,970 deer

Find the balance in the account.
2. $3,300 principal earning 4%, compounded annually, after 3 years
a. $3,712.05
b. $211,200.00
c. $3,696.00
d. $10,296.00

3. $1,600 principal earning 7%, compounded semi-annually, after 33 years
a. $4,979.11
b. $14,920.54
c. $112,992.00
d. $15,494.70

Answer 1

ok what do you think we have to do first?

Answer 2

I don't know... @hysusonic

Answer 3

ok for the first one, you have to first find how much 3% of 19,900 is

Answer 4

Okay so how do we do that?

Answer 5

multiply 19,900 with 0.03

Answer 6

597

Answer 7

@hysusonic

Answer 8

yup

Answer 9

now you multiply that by 10

Answer 10

5,970

Answer 11

Okay now add that to 19,900

Answer 12

25,870

Answer 13

So B?

Answer 14

No i messed up somewhere hold on a moment

Answer 15

The formula should follow this pattern: p = p(i) * r^n where p(n) is the population after n years, p(i) is the initial population, r is the rate of growth, and n is the number of years after the initial population was recorded.

So, the formula for calculating the population for n number of years is:

p(n) = 19,900 * 1.03^n

Because we want to predict the population after 10 years, we should substitute 10 for n in the formula.

p(10) = 19,900 * 1.03^10

Now, we simplify.

p(10) = 19,900 * 1.3439
p(10) = 26743.935

Because you can't have a fraction of deer, we must round the answer down. So, the deer population after 10 years is 26,743.

Answer 16

I need to hurry! what are the next 2?

Answer 17

Please :)

Answer 18

Next, using compounding: \[3300*1.04^3=3,712\]

Answer 19

Thanks you. How about 3? please

Answer 20

i'm not 100% sure but i think it's \[1600*1.07^33\]

Answer 21

1.07^33*

Answer 22

Thanks I got it.